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Unmasking Hidden Quadratics

This lesson explores how to solve quadratics through factorisation and how substitution reveals hidden quadratic structures within higher-degree equations.

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Factorising Quadratics
x2−5x+6=(x−2)(x−3)x^2 - 5x + 6 = (x-2)(x-3)
To solve x squared minus 5x plus 6 equals 0, we look for two numbers that multiply to 6 and add to negative 5. These numbers, negative 2 and negative 3, allow us to rewrite the expression as a product.
Step-by-step solver
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(a) Factorise

Identify two numbers that multiply to 6 and add to -5, which are -2 and -3.

x2−5x+6=(x−2)(x−3)=0x^2 - 5x + 6 = (x-2)(x-3) = 0
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(a) Solve

Set each bracket to zero to find the roots.

x−2=0  ⟹  x=2;x−3=0  ⟹  x=3x-2=0 \implies x=2; \quad x-3=0 \implies x=3
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(b) Substitution

Let y = x^2 to transform the quartic equation into a quadratic in y.

y2−5y+6=0  ⟹  y=2,y=3y^2 - 5y + 6 = 0 \implies y=2, y=3
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(b) Back-substitution

Solve x^2 = 2 and x^2 = 3 for x, considering both roots.

x=±2,x=±3x = \pm\sqrt{2}, \quad x = \pm\sqrt{3}

Original question

(a) Solve the quadratic equation x^2 - 5x + 6 = 0. (b) Hence solve the equation x^4 - 5x^2 + 6 = 0, giving all real roots.

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